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Thread: Represent inverse trigonometric equation as rational integral equation?

  1. #1
    Super Member fardeen_gen's Avatar
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    Represent inverse trigonometric equation as rational integral equation?

    Prove that the equation:
    $\displaystyle \cot^{-1}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) = 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}$
    can be represented as a rational integral equation in x & y.

    NOTE: No latex for arccot?
    EDIT: @Mush:
    Code:
    \acot
    doesn't work!
    EDIT: Thanks Soroban! Also, $\displaystyle \text{arccot}$ is the only way it seems.
    Last edited by fardeen_gen; Apr 28th 2009 at 06:49 AM. Reason: Follow up comments.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Prove that the equation:
    $\displaystyle \cot^{-1}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) = 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}$
    can be represented as a rational integral equation in x & y.

    NOTE: No latex for arccot?
    \acot
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  3. #3
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    Hello, fardeen_gen!

    Prove that the equation: .$\displaystyle \text{arccot}\left(\frac{y}{\sqrt{1 - x^2 - y^2}}\right) \;=\; 2\arctan\sqrt{\frac{3 - 4x^2}{4x^2}} - \arctan\sqrt{\frac{3 - 4x^2}{x^2}}$

    can be represented as a rational integral equation in $\displaystyle x$ and $\displaystyle y.$

    $\displaystyle \text{We have: }\;\arctan\underbrace{\left(\frac{\sqrt{1 - x^2 - y^2}}{y}\right)}_{\theta} \:= \:2\cdot\underbrace{\arctan\sqrt{\frac{3 - 4x^2}{4x^2}}}_{\alpha} - \underbrace{\arctan\sqrt{\frac{3 - 4x^2}{x^2}}}_{\beta}$

    . . That is: .$\displaystyle \theta \;=\;2\alpha - \beta$


    Take the tangent of both sides: .$\displaystyle \tan\theta \;=\;\tan(2\alpha + \beta)$

    . . $\displaystyle \tan\theta \;=\;\frac{\tan(2\alpha) + \tan\beta} {1 - \tan(2\alpha)\tan\beta} \;=\;
    \frac{\dfrac{2\tan\alpha}{1 - \tan^2\!\alpha} + \tan\beta}{1 - \dfrac{2\tan\alpha}{1 - \tan^2\!\alpha}\tan B} $

    . . $\displaystyle \tan\theta \;=\;\frac{2\tan\alpha - \tan\beta(1 - \tan^2\!\alpha)} {1-\tan^2\!\alpha - 2\tan\alpha\tan\beta}$

    . . $\displaystyle \tan\theta - \tan^2\!\alpha\tan\theta - 2\tan\alpha\tan\beta\tan\theta \;=\;2\tan\alpha - \tan\beta + \tan^2\!\alpha\tan\beta $ .[1]



    Since $\displaystyle \tan\theta = \frac{\sqrt{1-x^2-y^2}}{y},\quad \tan\alpha = \frac{\sqrt{3-4x^2}}{2x},\quad \tan\beta = \frac{\sqrt{3-4x^2}}{x} $

    substitute into [1]:

    . . $\displaystyle \frac{\sqrt{1-x^2-y^2}}{y} \;-\; \frac{3-4x^2}{4x^2} \cdot \frac{\sqrt{1-x^2-y^2}}{y} \;-\; 2 \cdot \frac{\sqrt{3-4x^2}}{2x}\cdot\frac{\sqrt{3-4x^2}}{x}\cdot\frac{\sqrt{1-x^2-y^2}}{y} $

    . . . . $\displaystyle = \;2\,\frac{\sqrt{3-4x^2}}{2x} - \frac{\sqrt{3-4x^2}}{x} + \frac{3-4x^2}{4x^2}\cdot\frac{\sqrt{3-4x^2}}{x} $


    Simplify: .$\displaystyle \frac{\sqrt{1-x^2-y^2}}{y} - \frac{5(3-4x^2)\sqrt{1-x^2-y^2}}{4x^2y} \;=\;\frac{(3-4x^2)^{\frac{3}{2}}}{4x^3} $


    Multiply by $\displaystyle 4x^3y\!:\quad 4x^3\sqrt{1-x^2-y^2} - 5x(3-4x^2)\sqrt{1-x^2-y^2} \;=\;y(3-4x^2)^{\frac{3}{2}} $


    Factor and simplify: .$\displaystyle 3x(8x^2-5)\sqrt{1-x^2-y^2} \;=\;y(3-4x^2)^{\frac{3}{2}} $


    Square both sides: .$\displaystyle 9x^2(8x^2-5)^2(1-x^2-y^2) \;=\;y^2(3-4x^2)^3$

    . . and we have a polynomial equaton in $\displaystyle x$ and $\displaystyle y.$

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